If int frac{2x-1}{(x-1)(x+2)(x-3)} dx = A log | vedclass

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(B) To find $A, B, C$,we use partial fraction decomposition: $\frac{2x-1}{(x-1)(x+2)(x-3)} = \frac{A}{x-1} + \frac{B}{x+2} + \frac{C}{x-3}$ Multiplyin

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$\frac{-1}{6}, \frac{-1}{3}, \frac{1}{2}$

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(B) To find $A, B, C$,we use partial fraction decomposition: $\frac{2x-1}{(x-1)(x+2)(x-3)} = \frac{A}{x-1} + \frac{B}{x+2} + \frac{C}{x-3}$ Multiplying both sides by $(x-1)(x+2)(x-3)$,we get: $2x-1 = A(x+2)(x-3) + B(x-1)(x-3) + C(x-1)(x+2)$ For $x=1$: $2(1)-1 = A(1+2)(1-3) \implies 1 = A(3)(-2) \implies A = -\frac{1}{6}$ For $x=-2$: $2(-2)-1 = B(-2-1)(-2-3) \implies -5 = B(-3)(-5) \implies -5 = 15B \implies B = -\frac{1}{3}$ For $x=3$: $2(3)-1 = C(3-1)(3+2) \implies 5 = C(2)(5) \implies 5 = 10C \implies C = \frac{1}{2}$ Thus,$A = -\frac{1}{6}, B = -\frac{1}{3}, C = \frac{1}{2}$.

If $\int \frac{2x-1}{(x-1)(x+2)(x-3)} dx = A \log |x-1| + B \log |x+2| + C \log |x-3| + K$,then $A, B, C$ are respectively:

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